## Calculus: Early Transcendentals 8th Edition

$$f'(x)=e^x(\sqrt x+\frac{1}{2\sqrt x})$$ $$f''(x)=e^x(\frac{1}{\sqrt x}+\sqrt x-\frac{1}{4x\sqrt x})$$
$$f(x)=\sqrt xe^x$$ $$f(x)=x^{1/2}e^x$$ 1) Find $f'(x)$ Using the Product Rule, we have $$f'(x)=x^{1/2}\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(x^{1/2})$$ $$f'(x)=x^{1/2}e^x+e^x\frac{1}{2}x^{-1/2}$$ $$f'(x)=e^x(x^{1/2}+\frac{1}{2}x^{-1/2})$$ $$f'(x)=e^x(\sqrt x+\frac{1}{2\sqrt x})$$ 2) Find $f''(x)$ We would start with $$f'(x)=e^x(x^{1/2}+\frac{1}{2}x^{-1/2})$$ Using the Product Rule again, we have $$f''(x)=e^x\frac{d}{dx}(x^{1/2}+\frac{1}{2}x^{-1/2})+(x^{1/2}+\frac{1}{2}x^{-1/2})\frac{d}{dx}(e^x)$$ $$f''(x)=e^x(\frac{1}{2}x^{-1/2}+\frac{1}{2}\times(-\frac{1}{2}x^{-3/2}))+(x^{1/2}+\frac{1}{2}x^{-1/2})e^x$$ $$f''(x)=e^x(\frac{1}{2}x^{-1/2}-\frac{1}{4}x^{-3/2})+e^x(x^{1/2}+\frac{1}{2}x^{-1/2})$$ $$f''(x)=e^x(x^{-1/2}+x^{1/2}-\frac{1}{4}x^{-3/2})$$ $$f''(x)=e^x(\frac{1}{\sqrt x}+\sqrt x-\frac{1}{4x\sqrt x})$$